Question

A simple pendulum is taken from earth to a place
where the value of acceleration due to gravity is g/4,
where g is the acceleration due to gravity on earth.
Calculate the time period of the pendulum in that
place.​

Answer 1

Answer:

Time period becomes twice as that on earth surface

Explanation:

$\large \: t = 2\pi \sqrt{ \frac{l}{g} } \: at \: earth$

$\huge \: {t}^{ \star} = 2\pi \sqrt{ \frac{2}{ \frac{g}{4} } }$

.

Hence

$2t = {t}^{ \star}$

$\small \: NOTE \: \: ({t}^{ \star} = time \: period \: above \: earth)$